Question

Evaluate $\int \frac{x^2+1}{x^4-x^2+1}dx$

• A. ${\tan }^{-1}\left(\frac{x^2+1}{x}\right)+c$
• B. ${\cot }^{-1}\left(\frac{x^2+1}{x}\right)+c$
• C. ${\cot }^{-1}\left(\frac{x^2-1}{x}\right)+c$
• D. ${\tan }^{-1}\left(\frac{x^2-1}{x}\right)+c$

Answer 1

The correct option is D ${\tan }^{-1}\left(\frac{x^2-1}{x}\right)+c$

Let $I=\int \frac{x^2+1}{x^4-x^2+1}dx=\int \frac{1+\frac{1}{x^2}}{x^2-1+\frac{1}{x^2}}dx=\int \frac{1+\frac{1}{x^2}}{{(x-\frac{1}{x})}^2+1}dx$
Put $x-\frac{1}{x}=t$

$\Rightarrow (1+\frac{1}{x^2})dx=dt$
$I=\int \frac{dt}{t^2+1}$

$={\tan }^{-1}t+c$ ..................... $\int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$={\tan }^{-1}(x-\frac{1}{x})+c$